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c4 integrating ln(x+2) by parts

integrate ln(x+2) by parts
my working: u=lnx+2 du/dx =1/x+2, dv/dx=1 v=1 woops meant x
this makes
xln(x+2)-integral of x/x+2
i got xln(x+2)-x+0.25x²+c

however in the mark scheme it says this

(x+2) \ln(x+2) -x + k

is my answer the same? and if it is would i be awared the full marks

(edited 11 years ago)

Reply 1

Enginerd.

Original post by printergirl

integrate ln(x+2) by parts
my working: u=lnx+2 du/dx =1/x+2, dv/dx=1 v=1
this makes
xln(x+2)-integral of x/x+2
i got xln(x+2)-x+0.25x²+c

however in the mark scheme it says this

(x+2) \ln(x+2) -x + k

is my answer the same? and if it is would i be awared the full marks

"dv/dx=1 v=1"

This is wrong. If DV/DX = 1, v = X.

Reply 2

printergirl

Original post by Enginerd.

"dv/dx=1 v=1"

This is wrong. If DV/DX = 1, v = X.

thats a typo :colondollar:

i used v=x in my workings

Reply 3

printergirl

also can someone help me with this question
5ii, how did they get root 13
http://www.thestudentroom.co.uk/attachment.php?attachmentid=159039&d=1340294617

Reply 4

GeneralOJB

Did you seriously divide x/x+2 into x/x + x/2? :eek:

Use substitution u=x+2

Reply 5

TenOfThem

Original post by printergirl

integrate ln(x+2) by parts
my working: u=lnx+2 du/dx =1/x+2, dv/dx=1 v=1 woops meant x
this makes
xln(x+2)-integral of x/x+2
i got xln(x+2)-x+0.25x²+c

however in the mark scheme it says this

(x+2) \ln(x+2) -x + k

is my answer the same? and if it is would i be awared the full marks

Why would

\int \dfrac{x}{x+2} dx

-x + \dfrac{x^2}{4}

Reply 6

printergirl

Original post by GeneralOJB

Did you seriously divide x/x+2 into x/x + x/2? :eek:

Use substitution u=x+2

can you not do that?? edit oh i c :colondollar:

best made now than in the exam

can any1 do question 5ii, how did they get root 13
http://www.thestudentroom.co.uk/atta...9&d=1340294617

how do you split x/x+2

(edited 11 years ago)

Reply 7

TenOfThem